Question

Check whether $\left( {5, - 2} \right),\left( {6,4} \right)$ and $\left( {7, - 2} \right)$ are the vertices of an isosceles triangle.

Answer 1

Hint: From the given question, we have to check whether the given vertices represent the isosceles triangle. We have to show that by using the distance formula between the vertices and check whether it contains a property of the isosceles triangle.
The distance between two points is a basic concept in geometry. It is so easy to learn the concept and work out. There are so many problems based on the distance concept to solve graphically.

Formula used: The distance formula between the two points $\left( {{{\text{x}}_{\text{1}}}{\text{,}}{{\text{y}}_{\text{1}}}} \right)$ and $\left( {{{\text{x}}_2}{\text{,}}{{\text{y}}_2}} \right)$ is given by
${\text{Distance}},{\text{d}} = \sqrt {{{\left( {{{\text{x}}_{\text{2}}} - {{\text{x}}_{\text{1}}}} \right)}^2} + {{\left( {{{\text{y}}_{\text{2}}} - {{\text{y}}_{\text{1}}}} \right)}^2}} $

Complete step-by-step solution:
Let us consider the given three vertices \[{\text{A}}\left( {5, - 2} \right),{\text{B}}\left( {6,4} \right)\] and ${\text{C}}\left( {7, - 2} \right)$ of the triangle ${\text{ABC}}$ respectively.
The figure of the given triangle ${\text{ABC}}$ in the following below:

Now, we have to check whether the given above vertices represent the isosceles triangle. First, we have to find the distance between every two points for ${\text{AB,BC}}$ and ${\text{CA}}$.
The distance between the two vertices \[{\text{A}}\left( {5, - 2} \right)\] and \[{\text{B}}\left( {6,4} \right)\] are
${\text{Distance}},{\text{d}} = \sqrt {{{\left( {{{\text{x}}_{\text{2}}} - {{\text{x}}_{\text{1}}}} \right)}^2} + {{\left( {{{\text{y}}_{\text{2}}} - {{\text{y}}_{\text{1}}}} \right)}^2}} $
\[{\text{AB}} = \sqrt {{{\left( {6 - 5} \right)}^2} + {{\left( {4 - \left( { - 2} \right)} \right)}^2}} \]
Simplifying we get,
${\text{AB}} = \sqrt {{{\left( {6 - 5} \right)}^2} + {{\left( {4 + 2} \right)}^2}} $
Add and subtract the terms,
${\text{AB}} = \sqrt {{{\left( 1 \right)}^2} + {{\left( 6 \right)}^2}} = \sqrt {1 + 36} = \sqrt {37} .$
The distance between the two vertices \[{\text{B}}\left( {6,4} \right)\] and ${\text{C}}\left( {7, - 2} \right)$ are
${\text{Distance}},{\text{d}} = \sqrt {{{\left( {{{\text{x}}_{\text{2}}} - {{\text{x}}_{\text{1}}}} \right)}^2} + {{\left( {{{\text{y}}_{\text{2}}} - {{\text{y}}_{\text{1}}}} \right)}^2}} $
Simplifying we get,
${\text{BC}} = \sqrt {{{\left( {7 - 6} \right)}^2} + {{\left( { - 2 - 4} \right)}^2}} $
Add and subtract the terms,
${\text{BC}} = \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 6} \right)}^2}} $
Hence,
${\text{BC}} = \sqrt {{{\left( 1 \right)}^2} + {{\left( 6 \right)}^2}} = \sqrt {1 + 36} = \sqrt {37} .$
The distance between the two vertices ${\text{C}}\left( {7, - 2} \right)$ and \[{\text{A}}\left( {5, - 2} \right)\] are
${\text{Distance}},{\text{d}} = \sqrt {{{\left( {{{\text{x}}_{\text{2}}} - {{\text{x}}_{\text{1}}}} \right)}^2} + {{\left( {{{\text{y}}_{\text{2}}} - {{\text{y}}_{\text{1}}}} \right)}^2}} $
Substituting the values,
${\text{CA}} = \sqrt {{{\left( {5 - 7} \right)}^2} + {{\left( { - 2 - \left( { - 2} \right)} \right)}^2}} $
Add and subtract the terms,
${\text{CA}} = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2 + 2} \right)}^2}} $
Hence,
${\text{CA}} = \sqrt {{{\left( 2 \right)}^2} + {{\left( 0 \right)}^2}} = \sqrt {4 + 0} = \sqrt 4 = 2.$
Here, we found the distance for the three vertices which show that the distance between ${\text{AB}}$ and \[{\text{BC}}\] sides having the lengths are equal.
By the property of the isosceles triangle, thus we have the two sides are equal, ${\text{AB}} = {\text{BC}} = \sqrt {37} $.

$\therefore $ The given vertex forms an isosceles triangle.

Note: We can observe that from the solution of the question, the distance formula has been derived for two points which are not on a horizontal line or vertical line.
i) Three given points are collinear or form a right triangle, isosceles triangle or equilateral triangle.
ii) Four given points form a parallelogram, rectangle, square or rhombus.
An isosceles triangle is a triangle that has (at least) two equal side lengths. If all three side lengths are equal, the triangle is also equilateral. Isosceles triangle is very helpful in determining unknown angles.